A215494 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3) with a(1)=7, a(2)=21, a(3)=70.

7, 21, 70, 245, 882, 3234, 12005, 44933, 169099, 638666, 2417807, 9167018, 34790490, 132119827, 501941055, 1907443237, 7249766678, 27557748813, 104759610858, 398257159370, 1514069805269, 5756205681709, 21884262613787, 83201447389466, 316323894905207

Offset: 1

Author

Roman Witula, Aug 13 2012

Keywords

• nonn
• easy

Comments

The Berndt-type sequence number 5 for the argument 2*Pi/7; see also A215007, A215008, A215143, A215493 and A215510.

We note that if we set:

x(n) := s(2)*s(1)^n + s(4)*s(2)^n + s(1)*s(4)^n,

y(n) := s(4)*s(1)^n + s(1)*s(2)^n + s(2)*s(4)^n,

z(n) := s(1)^(n+1) + s(2)^(n+1) + s(4)^(n+1),

for every n=0,1,..., where s(j) := 2*sin(2*Pi*j/7), then the following system of recurrence equations holds true:

x(n+2)=2*x(n)-y(n), y(n+2)=2*y(n)-x(n)+z(n), z(n+2)=y(n)+3*z(n).

Moreover we have a(n)=z(2*n-1), A215493(n)=z(2*n-2), A094429(n)=y(2n-1)-x(2n-1)=-x(2*n+2)/sqrt(7), A094430(n)=-x(2*n+3), y(2*n-2)=sqrt(7)*A215143(n), y(2*n-1)=A215510(n) and x(11)=-(y(10)+z(10))/sqrt(7)=-1078.

We can also deduce the following relations:

x(n-1) = c(1)*s(1)^n + c(2)*s(2)^n + c(4)*s(4)^n,

-y(n-1)-z(n-1) = c(2)*s(1)^n + c(4)*s(2)^n + c(1)*s(4)^n,

y(n-1)-x(n-1) = c(4)*s(1)^n + c(1)*s(2)^n + c(2)*s(4)^n,

for every n=1,2,..., where x(0)=y(0)=z(0)=sqrt(7), and c(j) := 2*cos(2*Pi*j/7).

All these sequences satisfy the following recurrence equation: Z(n+6)-7*Z(n+4)+14*Z(n+2)-7*Z(n)=0. The characteristic polynomial of this equation (after rescaling) has the form (X-s(1)^2)*(X-s(2)^2)*(X-s(3)^2)=X^3-7*X^2+14*X-7 and was recognized by Johannes Kepler (1571-1630); see the Savio-Suryanarayan paper.

We also have the following decomposition: (X-s(1)^(n+1))*(X-s(2)^(n+1))*(X-s(4)^(n+1)) = X^3 - z(n)*X^2 + (1/2)*(z(n)^2-z(2n+1))*X - (-sqrt(7))^(n+1).

Further we have a(n)=A146533(n) for n=1,...,6, and A146533(7)-a(7)=7. We note that all numbers 7^(-1-floor(n/3))*a(n) are integers.

Examples

			We have a(3)=5*7^2 and a(6)=5*7^4, which implies that s(1)^12 + s(2)^12 + s(4)^12 = 49*(s(1)^6 + s(2)^6 + s(4)^6). We also have a(9) = (a(1) + a(3))*7^49.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• B. C. Berndt, A. Zaharescu, Finite trigonometric sums and class numbers, Math. Ann. 330 (2004), 551-575.
• B. C. Berndt, L.-C. Zhang, Ramanujan's identities for eta-functions, Math. Ann. 292 (1992), 561-573.
• L. Bankoff and J. Garfunkel, The Heptagonal Triangle, Math. Magazine, 46 (1973), 7-19.
• Z.-G. Liu, Some Eisenstein series identities related to modular equations of the seventh order, Pacific J. Math. 209 (2003), 103-130.
• D. Y. Savio and E. R. Suryanarayan, Chebyshev polynomials and regular polygons, Amer. Math. Monthly, 100 (1993), 657-661.
• Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
• Index entries for linear recurrences with constant coefficients, signature (7, -14, 7).

Crossrefs

See A122068.

Programs

Magma

I:=[7,21,70]; [n le 3 select I[n] else 7*Self(n-1)-14*Self(n-2)+7*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 01 2016

Mathematica

LinearRecurrence[{7,-14,7}, {7,21,70}, 50]

PARI

polsym(x^3 - 7*x^2 + 14*x - 7, 30) \\ (includes a(0)=3) Joerg Arndt, May 31 2017

PARI

x='x+O('x^30); Vec((7-28*x+21*x^2)/(1-7*x+14*x^2-7*x^3)) \\ G. C. Greubel, Apr 23 2018

Formula

Equals 7*A122068. - M. F. Hasler, Aug 25 2012

a(n) = s(1)^(2n) + s(2)^(2n) + s(4)^(2n), where s(j) := 2*Sin(2*Pi*j/7) (for the sums of the respective odd powers see A215493, see also A094429, A115146). For the proof of these formula see Witula-Slota's paper.

G.f.: (7 - 28*x + 21*x^2)/(1 - 7*x + 14*x^2 - 7*x^3) = -d(log(1 - 7*x + 14*x^2 - 7*x^3))/dx.